Question
Derive an expression for the excess pressure inside a liquid drop.
✓
Answer
Consider a liquid drop of radius r. If one tries to enhance the radius by a small amount ‘dr’ work has to be done to overcome the excess pressure (P). 
Work done $=\text{dW}=\text{p}\times4\pi\text{r}^2\times\text{dr}$ Due to surface tension $'\sigma',$ the excess pressure exists. The work done to change the area is also written as,$\text{dW}=\sigma \times\text{change in area}$
$=\sigma\times4\pi\{(\text{r}+\text{dr})^2-\text{r}^2\}=\sigma8\pi\text{rdr}$
$\therefore\text{P}4\pi\text{r}^2\text{dr}=\sigma8\pi\text{r}\text{ dr}$
$\therefore\text{P}=\frac{2\sigma}{\text{r}}$
Work done $=\text{dW}=\text{p}\times4\pi\text{r}^2\times\text{dr}$ Due to surface tension $'\sigma',$ the excess pressure exists. The work done to change the area is also written as,$\text{dW}=\sigma \times\text{change in area}$$=\sigma\times4\pi\{(\text{r}+\text{dr})^2-\text{r}^2\}=\sigma8\pi\text{rdr}$
$\therefore\text{P}4\pi\text{r}^2\text{dr}=\sigma8\pi\text{r}\text{ dr}$
$\therefore\text{P}=\frac{2\sigma}{\text{r}}$
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